MCQ: Clock and Calendar - 1 - SSC CGL MCQ

The year 2004 is a leap year. So, it has 2 odd days.
Feb 2004 is not included because we are calculating from April 2004 to April 2005.
Thus, it has 1 odd day only.
8th April, 2004 will be 1 day before the day on 8th April, 2005.
Given that, 8th March, 2005 is Monday.
So 8th March 2004 is Sunday (1 day before to 8th March, 2005).

Total hours from Monday 10.00 AM to Friday 3.00 PM =101 Hours.
The clock loses 6 minutes every hour. In 101 hours it will lose 101 x 6 =606 minutes. i.e. 10 hours 06 minutes.
To find the actual time, calculate 10 hours 6 min backwards from Friday 3.00 PM which is 4 .54 AM
Hence, the actual time would be 4 .54 AM when clock shows 3.00 PM on Friday.

Count the number of odd days from the year 2007 onwards to get the sum equal to 0 odd day.

Sum = 14 odd days 0 odd days.
Years 2018 and 2007 will the same calendar.
Option 4 is the correct answer.

55 min. spaces are covered in 60 min
60 min. spaces are covered
in ((60/55) x 60) min = 720/11 min.
in 64 min. the loss is = (720/11 - 64). = 16/11
In 24 hrs the loss is = (16/11 x 1/64 x 24 x 60) min = 32 x 8/11 = 256/11
Option 1 is the correct answer.

57 after 12 o’clock 12:21……= 1 from 1 to 9 it is 1.01, 1.11, 1.21, 1.31, 1.41, 1.51 = 6 similarly 2.02, 2.12… 6*9 = 54+1 = 55 after 10 o’clock 10:01, 11:11 Ans: 55 + 2 = 57
Option 3 is the correct answer.

2/7 in a leap year there are 366 days means 52 weeks and 2days. So already we have 52 Fridays. Now the rest two days can be (Sunday, Monday), (Monday, Tuesday), (Tuesday, wed), (Wednesday, Thursdays), (Thursday, Friday), (Friday, sat), (sat, sun) so, the probability of 53 Fridays= 2/7
Option 3 is the correct answer

An hours + C hours = B hours … (I)
A, C and B can’t possess values greater than or equal to 24 B minutes + A minutes = C minutes … (ii)
From the above two equations, it is assumed that no value of A satisfies both the equations.
Hence, option 1 is the correct answer.

This time period was 1998 which was not a leap year. Now, we calculate the no. of odd days in 1999 up to February 19th
January 1999 gives 3 odd days
19 February 1999 gives 5 odd days
1998, being ordinary year, gives 1 odd day
1997 December 30th and 31th have 2 odd days
The totalling of number of odd days = 3 + 5 + 1 + 2 = 11 days = 4 odd days
Hence December 30, 1997 will fall 4 days prior Saturday i.e. on Tuesday.
Option 2 is the correct answer.

The clock loses 1% time during the first week.
A day comprises of 24 hours and a week has 7 days. Therefore, there are 7 * 24 = 168 hours in a week.
During the first week the clock loses 1% time, then it will show a time which is 1% of 168 hours less than 12 Noon at the end of the first week = 1.68 hours less.
The clock gains 2% during the next week subsequently. The clock gains 2% time in the second week as it has 168 hours = 2% of 168 hours = 3.36 hours more than the actual time.
It had lost 1.68 hrs during the first week and then again gained 3.36 hrs during the next week, the net result will be a -1.68 + 3.36 = 1.68 hour net gain in time.
Hence the clock will display time which is 1.68 hours more than 12 Noon two weeks from the time it was set.
1.68 hrs = 1 hr and 40.8 min = 1 hour + 40 min + 48 sec.
I.e. 1: 40: 48 P.M. Option 2 is the correct answer.

After every 400 years, the same day occurs. Before 400 years if 27th February 2003 is Thursday then on 27th February 1603 has to be Thursday.
Option 2 is the correct answer.

Normal year jump +1 gain; Leap year jump +2 gain.

Sunday plus five days
Option 2 is the correct answer

Let’s assume the race started at 5 hours X minutes and ended at 8 hours Y minutes. Angles made by both the hour’s hands are 150+X/2 and 240+Y/2. Angles made by the minutes hands are 6X and 6Y.

We know that the difference between the angles made by the hands of the clock is same, so we equate that: 6X – (150 + X/2) = (240 + Y/2) – 6Y => 12X – 300 – X = 480 + Y – 12 Y => 11X + 11Y = 480+300 => X+Y = (780/11) => X+Y ≈ ≈ 71

We know that the sum of the angles made by the hands of the clock is also same, so we equate that as well: 6X+150+X/2 = 240+Y/2+6Y => 13X + 300 = 480 + 13Y => 13(X-Y) = 180 => X-Y = (180/13) => X-Y = ≈ ≈ 14

Solving we two equations, we get X = 42.5 mins = 42 mins 30 secs and Y = 28.5 mins = 28 mins 30 secs

Hence, duration of the race is = 8:28-5:42 = 2 hours 46 mins

Option 2 is the correct answer

At any time, A hours and B minutes, the angle made by the hours hand is 30 A + B / 2. 30A+B/2. Similarly, the angle made by the minutes hand is 6B.

For a person to not be able to find the correct time, there should be another time     A1:B1 such that 30 A1 + B1 / 2 = 6B and 6B1 = 30 A + B / 2 6B1=30A+B/2

There is no integral solution existing when 0 < B < 60

Option 4 is the correct answer.

Second hand will move 360 degree in one minutes. Hence time is = 3960/360 = 11 minutes so hr hand will travel 11*30/60 = 5.5 degree.
Option 2 is the correct answer.

730/7 = 2 odd days. Hence Saturday will be the day of the week.
Option 3 is the correct answer.